Calculating sample size required for a given width and st. dev.

I’ve come across a discrepancy in one of my practice questions.

I’m given the following: σ=150\sigma=150 units. For a 98% CI, how many observations do I need to be accurate within 15 units.

Now, the 15 refers to the width of my CI: w=2⋅Zα2⋅σ√nw=2\cdot Z_{\frac{\alpha}{2}}\cdot \frac{\sigma}{\sqrt{n}}

This comes from w=(ˉx+Zα2⋅σ√n)−(ˉx−Zα2⋅σ√n)w=(\bar{x}+Z_{\frac{\alpha}{2}}\cdot \frac{\sigma}{\sqrt{n}})-(\bar{x}-Z_{\frac{\alpha}{2}}\cdot \frac{\sigma}{\sqrt{n}})

For a 98% CI, 2-tailed, I need the Z-score for 0.99. That is 2.33 appoximately (qnorm(.99)=2.3263).

so when I plug it in and solve for n:
n=(2⋅2.33⋅150/30)2n=(2\cdot 2.33\cdot 150/30)^2 =23.32=543=23.3^2=543.

However, the answer given is 644…. if someone could help me out I would appreciate it.




It appears as if 664 gives the answer for a 99% CI (i.e. find the Z for 99.5%)… is the solution incorrect?
– sahimat
Oct 20 at 20:58



Agreed. Your answer looks correct to me, if it’s a 98% CI
– Dr Xorile
Oct 20 at 21:54